In the rhombus ABCD, AB=10 and AC=12. Find the area of the rhombus.

I am confused about how to solve this problem.

let E be the intersection of the diagonals AC andBD

In a rhombus the diagonals right-bisect each other
so triangle ABE has
AB=12 , AE = 6 , and by Pythagoras BE = 8

Area of rhomus = 1/2 of the product of its diagonals
= (1/2)(12)(16) = 96

How do I know the value of the diagonals?

You told me one was 12 , AC = 12

and I found half of the other for you
I said BE = 8, so BD = 16

Ok, I understand that. But I am confused as to how you found the second diagonal (BD)?

To find the area of a rhombus, you can use the formula:

Area = (diagonal_1 * diagonal_2) / 2

In this case, the diagonals of the rhombus are not given directly. However, we can find them using the given information.

Let's start by drawing the rhombus ABCD:

A
/ \
/ \
/ \
/ \
B ------- C
\ /
\ /
\ /
\ /
D

We know that AB = 10 and AC = 12. Since a rhombus has equal sides, we can also say AD = BC = 10, and BD = AC = 12.

Now, we can observe that the diagonals of a rhombus intersect each other at right angles and bisect each other. Let's name the point where the diagonals intersect as E.

A
/ \
/ \
/ E \
/-------\
B ------- C
\ /
\ /
\ /
\ /
D

Since the diagonals bisect each other, we can say that AE = EC = 12/2 = 6 and BE = ED = 10/2 = 5.

Now, we have the lengths of the diagonals. We can substitute these values into the formula for the area of a rhombus:

Area = (diagonal_1 * diagonal_2) / 2
= (BE * AC) / 2
= (5 * 12) / 2
= 60 / 2
= 30 square units

So, the area of the rhombus ABCD is 30 square units.